In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

Would appreciate any help

thanks :)

Sep 12th 2008, 10:52 AM

topsquark

Quote:

Originally Posted by gracey

Could do with some help

In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

Would appreciate any help

thanks :)

There are four basic kinematics equations for motion under constant acceleration. You are going to use two of them here.

For the first part we know that
So set the zero level at the top of the volcano and upward as positive. Notice that at the top of the motion the speed is 0 m/s. So we get
which you can solve for .

For the second part we also know that
and again v = 0 m/s so
You know from the first part, so now just solve for t.

-Dan

Sep 12th 2008, 11:05 AM

arbolis

Just for fun, as (the gravitational force on the Earth's surface) is not constant over these , you can also find with the following formula : . Considering the Earth as a sphere with radius , and I found that is worth . Compare this result with yours, I think mine should be a bit lesser than yours.

Sep 12th 2008, 11:11 AM

Soroban

Hello, gracey!

Quote:

In a volcanic eruption, a rock is ejected vertically from the crater
and rises to a height of 25 kilometres.

Find the speed with which it emerges from the crater,
and time it takes to reach this height.

The equation is: .

where: . . measured in meters and seconds

Assume . . We will measure the rock's height from the top of the volcano.

The equation is: .

This is a down-opening parabola; its peak is at its vertex.. .

With and , we have: . seconds. .[1]

The maximum height is: . meters

Since the maximum height is 25 km = 25,000 m,. .

Therefore, the initial velocity is: .

Substitute into [1]: .

Sep 12th 2008, 11:17 AM

arbolis

What?! Soroban, an initial speed of almost ! And it takes to reach only ! You certainly mean which surprises me because I thought it would have been more than .
I don't see where is my error...

Sep 12th 2008, 11:36 AM

topsquark

Quote:

Originally Posted by Soroban

The maximum height is: . meters

Since the maximum height is 25 km = 25,000 m,. .

Therefore, the initial velocity is: .
There was a typo or something here.

Substitute into [1]: .

So
-Dan

Sep 12th 2008, 11:48 AM

arbolis

topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?

Sep 12th 2008, 12:17 PM

topsquark

Quote:

Originally Posted by arbolis

topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?

You are correct. I must have punched the numbers in wrong. I have fixed it in my response.

The point is that Soroban forgot to square the in his height equation. And notice that 490 km/s is significantly higher than escape velocity, so it won't have a max. height. :)